t2+4t+19=42

Solution for t2+4t+19=42 equation:

t2+4t+19=42

We move all terms to the left:

t2+4t+19-(42)=0

We add all the numbers together, and all the variables

t^2+4t-23=0

a = 1; b = 4; c = -23;
Δ = b2-4ac
Δ = 42-4·1·(-23)
Δ = 108
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{108}=\sqrt{36*3}=\sqrt{36}*\sqrt{3}=6\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-6\sqrt{3}}{2*1}=\frac{-4-6\sqrt{3}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+6\sqrt{3}}{2*1}=\frac{-4+6\sqrt{3}}{2}
$